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php使用json_encode对变量json编码

Jun 13, 2016 am 09:39 AM
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 这篇文章主要是说明 json_encode 使用时一个需要注意的细节,无论返回的是数组还是 json,都不代表这一定就是个错误的返回结果,当前端需要的数组时,数组就是正确的结果,反之亦然

在 ajax 横行的今天,json 顺理成章地成为了前后端交互一个极其重要的工具。对于 PHP 来说,通常使用 json_encode 方法将一个 PHP 组数,转换成前端可以解析的 json 字符串,这也是 PHP 手册上描述的内容,但事实是这样的吗?看看下面这段代码:

 

代码如下:

$a = array( 'Jack' , 'Sam' , 'Tom' );

echo json_encode( $a );

 

当 JavaScript 请求到上面的代码,PHP 将解析数组 $a 为 json 字符串并返回到前端,但事实上在前端拿到的返回结果却是一个数组。

代码如下:

[ "Jack" ,  "Sam" ,  "Tom" ]

 

这个结果对于前端来说也许并不是所期望的,对于 JavaScript 来说,数组和 json 的关系非常密切,甚至可以简单地把 json 理解为关联数组,但这并不意味着二者可以划等号,比如 json 没有 length 属性,也不能进行数字索引,json 是键值对,而 JavaScript 的数组严格说来是没有"键"的,这和 PHP 的差异比较大。那么为什么 json_encode 的返回结果是数组呢?

 

本文开头的 PHP 代码中的数组是严格意义上的数字索引数组,json_encode 方法在处理这样的数组的时候会返回数组字符串,它需要同时满足两个条件:1. 数字索引数组,2. 索引值从 0 开始。这意味着下面的代码也会返回同样的结果:

 

 

代码如下:

$b = array(

'0' => 'Jack',

'1' => 'Sam',

'2' => 'Tom'

);

echo json_encode( $b );

 

这两个条件的任意一个没有获得满足,那么 json_encode 方法才真正的返回 json 字符串:

代码如下:

$c = array(

'person-1' => 'Jack',

'person-2' => 'Sam',

'person-3' => 'Tom'

);

echo json_encode( $c );

 

这时前端得到的结果如下:

代码如下:

{

'person-1' : 'Jack',

'person-2' : 'Sam',

'person-3' : 'Tom'

};

 

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